1. Periodic Trends - Warmup
Use your periodic table to help write noble gas electron configurations for the following elements: Sodium (Na) Sulfur (S) Copper (Cu) Tin (Sn) Erbium (Er) Gold (Au) Element 118

2. Periodic Trends - Warmup
Use your periodic table to help you write noble gas electron configurations for the following elements: Sodium (Na) 11 – [Ne]3s1 Sulfur (S) 16 - [Ne]3s23p4 Copper (Cu) 29 – [Ar]4s23d9 Tin (Sn) 50 – [Kr]5s24d105p2 Erbium (Er) 68 – [Xe]6s24f12 Gold (Au) 79 - [Xe]6s24f145d9 Element 118 [Rn]7s15f146d107p6

3. Electron Dot Practice
Write electron dot diagrams for the following elements:
All of the elements in Period 3 (beginning with Na)

4. Electron Dot Practice
Write electron dot diagrams for the following elements:
All of the elements in Period 3 (beginning with Na)
Put 1 dot on each side, then double up
Na = 1 dot, Ar = 8 dots, 2 on each side

5. Warmup - Overview of PT
From memory, compare metals, metalloids, nonmetals and noble gases in terms of:
Appearance
State of matter at room temperature
Electrical conductivity
Heat conductivity
Malleability and ductility
Electron configuration
Then check your notes

6. Periodic Trends - Warmup
1. Compare the following elements in terms of:
- atomic radius
- ionization energy
- electronegativity
Explain your reasoning for atomic radius and ionization energy.
a) K and Li
b) Al and Ar
c) F and Ba

7. K and Li – Atomic Radius
K has a larger radius than Li. K is further down Group 1 than Li. This means K has more principal energy levels than Li. The electrons that make up the principal energy levels (core electrons) provide a shield for the valence electron from the nucleus, allowing the valence electron to be further away from the nucleus than Li’s valence electron, which has only one principal energy.

8. K and Li – Ionization Energy
Li has a higher ionization energy than K because Li’s valence electron is closer to its nucleus than K’s principal energy level. See notes above about shielding effect. Closer to the nucleus means more attraction of the valence electron to the nucleus, making it harder to remove it. Thus Li requires more energy to remove its valence electron than does K.
Li has a higher electronegativity than does K.

9. Al and Ar – Atomic Radius
Ar has a smaller radius than Al. Ar is further across Period 3 than Al. This means Ar has more protons than Al, but the same number of core electrons as Al (i.e. same amount of shielding). The larger number of protons in Ar overcome the shield to a greater extent than in Al. As a result the valence electrons of Ar are pulled in closer to the nucleus, resulting in a smaller atomic radius.

10. Al and Ar – Ionization Energy
Ar has a higher ionization energy than Al because Ar’s valence electron are closer to its nucleus. See notes above about shielding effect. Closer to the nucleus means more attraction of the valence electron(s) to the nucleus, making it harder to remove it (them). Thus Ar requires more energy to remove its valence electrons than does Al.
Ar has no electronegativity, so by default, Al has the higher electronegativity.

11. F and Ba – Atomic Radius
Ba has a larger radius than F. F is one of the smallest elements in the periodic table, as it is both near the top and to the right. Ba is one of the larger atoms, as it is down near the bottom of the periodic table, and towards the left.
See previous slides for explanations incorporating the shielding effect.

12. F and Ba – Ionization Energy
F has a higher ionization energy than Ba because F’s valence electrons are closer to its nucleus. See notes above about shielding effect. Closer to the nucleus means more attraction of the valence electron(s) to the nucleus, making it harder to remove it (them). Thus F requires more energy to remove its valence electrons than does Ba.
F has the highest electronegativity in the periodic table. Ba’s electronegativity is low.

13. Periodic Trends - Warmup
2. Ca2+ and K+ each have 18 electrons around the nucleus.
Which has a smaller radius and why?

14. Periodic Trends - Warmup
2. Ca2+ and K+ each have 18 electrons around the nucleus.
Which has a smaller radius and why?
Ca2+ has the smaller radius. Ca2+ and K+ have the same number of core electrons (18), but Ca2+ has one more proton. This means Ca is smaller than K. In addition, K only loses 1 electron to become an ion, halving its size. Ca loses 2 electrons, making it even smaller.

15. Periodic Trends Warmup
1. List in order of increasing atomic radius:
Be, C, Li+, Ne, B3+, F-
2. List in order of decreasing ionization energy:
Cs, Na, Ca, H+, K+, Fr

16. Periodic Trends Warmup
1. List in order of increasing atomic radius:
B3+, Ne, Li+, C, Be, F-
2. List in order of decreasing ionization energy:
  H+, K+, Ca, Na, Cs, Fr
 Note: You will not see questions this difficult on the test